document.write( "Question 583086: PLEASE help! My teacher didn't explain.
\n" ); document.write( "I have to write a rational function with the given asymptotes.\r
\n" ); document.write( "\n" ); document.write( "a. x=-2, y=0
\n" ); document.write( "b. x=4,y=0
\n" ); document.write( "c.x=2,x=1,y=1
\n" ); document.write( "d.x=0,y= -1\r
\n" ); document.write( "\n" ); document.write( "Thank you so much! \n" ); document.write( "
Algebra.Com's Answer #372343 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
a) Vertical asymptotes. like x=-2, happen only when a denominator is zero.
\n" ); document.write( "\"f%28x%29=1%2F%28x%2B2%29\" has a vertical asymptote at x=-2
\n" ); document.write( "As x approaches -2, the denominator approaches zero and the absolute value of f(x) grows without bounds. The graph of \"f%28x%29=1%2F%28x%2B2%29\" looks like this:
\n" ); document.write( "\"graph%28300%2C300%2C-5%2C1%2C-50%2C50%2C1%2F%28x%2B2%29%29\" Coincidentally that function also has \"y=0\" as an asymptote, as you can see from the graph.
\n" ); document.write( "As the absolute value of x (and consequently of x+2) grows larger, and larger, f(x) grows closer and closer to zero. A \"y=0\" horizontal asymptote happens when your rational function is a quotient and the denominator polynomial has a higher degree than the numerator.
\n" ); document.write( "CAUTION: Not every time a denominator is zero, you have a vertical asymptote.
\n" ); document.write( "If you make sure that the denominator, and only the denominator is zero at x=-2, you can be sure that the function will have an x=-2 asymptote.
\n" ); document.write( "If the numerator and denominator are zero at the same time, the function can be equivalent to another function that does not have a vertical asymptote.
\n" ); document.write( "For example, \"p%28x%29=%28x%2B2%29%2F%28x%2B2%29\" is equivalent to \"q%28x%29=1\" for all values of x except x=-2, and you know that \"q%28x%29=1\" graphs as a horizontal line with \"y=1\", and does not have a vertical asymptote. The graph for \"p%28x%29=%28x%2B2%29%2F%28x%2B2%29\" looks just like the same horizontal y=1 line, except for a hole at x=-2, where p(x) does not exist.
\n" ); document.write( "b) From what I said above, you must realize that for a vertical \"x=4\" asymptote, you need the denominator to be zero for \"x=4\".
\n" ); document.write( "\"g%28x%29=1%2F%28x-4%29\" would work. It also has a \"y=0\" asymptote, because the denominator, x-4 has degree 1, and the numerator, 1, has degree zero.
\n" ); document.write( "c) \"h%28x%29=1%2F%28%28x-1%29%28x-2%29%29\" has asymptotes \"x=1\" and \"x=2\" because those are zeros of the denominator.
\n" ); document.write( "The only horizontal asymptote for \"h%28x%29=1%2F%28%28x-1%29%28x-2%29%29\" is \"y=0\" and we need \"y=1\", but that is easy to fix: we just add 1.
\n" ); document.write( "\"m%28x%29=1%2Bh%28x%29=1%2B1%2F%28%28x-1%29%28x-2%29%29\" has \"x=1\", \"x=2\" and \"y=1\" asymptotes.
\n" ); document.write( "You can make it look fancier:
\n" ); document.write( "\"m%28x%29=1%2B1%2F%28%28x-1%29%28x-2%29%29\"=\"%28x-1%29%28x-2%29%2F%28%28x-1%29%28x-2%29%29\"+\"1%2F%28%28x-1%29%28x-2%29%29\"=\"%28x%5E2-3x%2B2%29%2F%28x%5E2-3x%2B2%29\"+\"1%2F%28x%5E2-3x%2B2%29\"=\"%28x%5E2-3x%2B3%29%2F%28x%5E2-3x%2B2%29\"
\n" ); document.write( "d) For an \"x=0\" asymptote we want x as a factor in the denominator, but not tin the numerator. A \"y=0\" asymptote would be easier, but you saw in part c) how you can get a horizontal asymptote at a different y value
\n" ); document.write( "\"r%28x%29=-1%2B1%2Fx\"=\"%281-x%29%2Fx\" has \"x=0\" and \"y=-1\" asymptotes.
\n" ); document.write( "\"s%28x%29=-1\"+\"1%2Fx%5E2\"=\"%281-x%5E2%29%2Fx%5E2\" would work too. \n" ); document.write( "
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